Question: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{2})^{-1}}}{{(y^{-1}r^{4})^{-4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{2}}$ to the exponent ${-1}$ . Now ${2 \times -1 = -2}$ , so ${(y^{2})^{-1} = y^{-2}}$ In the denominator, we can use the distributive property of exponents. ${(y^{-1}r^{4})^{-4} = (y^{-1})^{-4}(r^{4})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{2})^{-1}}}{{(y^{-1}r^{4})^{-4}}} = \dfrac{{y^{-2}}}{{y^{4}r^{-16}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{-2}}}{{y^{4}r^{-16}}} = \dfrac{{y^{-2}}}{{y^{4}}} \cdot \dfrac{{1}}{{r^{-16}}} = y^{{-2} - {4}} \cdot r^{- {(-16)}} = y^{-6}r^{16}$.